Use nested polynomial evaluation one coefficient at a time

Nested polynomial evaluation gives each coefficient a turn in a short multiply-and-add sequence. The input value is already known. Begin with the leading coefficient, multiply the running value by the input, then add the next coefficient. Repeat in descending-power order until the constant term has been included. A zero for a missing power must keep its place in that order.

This original worksheet evaluates 2x³ − 3x² + 5x + 4 at x equal to three. Its running values are two, three, fourteen and forty-six. The arithmetic below explains what each value represents, checks the answer directly and then changes the example to show the effect of an omitted zero coefficient.

Record the complete polynomial first

Write P(x) = 2x³ − 3x² + 5x + 4. The coefficients, read from the highest displayed power down to the constant term, are 2, −3, 5 and 4. The negative sign belongs to the coefficient −3. Preserve it in the list instead of treating the list as four unsigned magnitudes.

Beside the expression, record the given input x = 3. Here x is a value to substitute, not an unknown to find by solving an equation. The goal is the value P(3). None of these numbers is a measurement or a product specification. They are exact integers chosen for this abstract example.

Associate every coefficient with its power before beginning: two with the cube, negative three with the square, five with the first power and four with power zero. This initial alignment prevents an otherwise orderly sequence from evaluating the wrong expression. An arithmetic check can pass while the starting coefficient list still misstates the intended polynomial.

Start with the leading coefficient

Set the running value to two, the first coefficient. No multiplication has happened yet. The remaining coefficients are −3, 5 and 4, in that order. Keep the input three visible beside the working, because the same input is used at every multiplication. The running value changes; the chosen input stays fixed.

For the first update, multiply two by three to obtain six. Then add negative three, giving three. Write the complete line as 2 × 3 − 3 = 3. The first three in this equation is the input; the final three is the updated running value. Their equality in this example does not merge their roles.

For the second update, multiply the running three by the input three to obtain nine, then add the next coefficient five. The new running value is fourteen. The complete line is 3 × 3 + 5 = 14. This update consumes the coefficient of the first-power term, leaving only the constant four to include.

For the final update, multiply fourteen by three to obtain forty-two, then add four. The result is forty-six. Write 14 × 3 + 4 = 46 and mark that the coefficient list is exhausted. The final running value is P(3), because every coefficient has now entered at its specified stage.

Expand the nested expression to check its meaning

The three updates can be written as one expression: ((2 × 3 − 3) × 3 + 5) × 3 + 4. The inner parentheses establish the first result before it is multiplied again. Reading the operations from the inside outward reproduces the recorded running values rather than asking the reader to infer an unstated order.

Before substituting three, the same structure is ((2x − 3)x + 5)x + 4. Expanding the first two stages gives 2x² − 3x + 5. Multiplying that expression by x and adding four gives 2x³ − 3x² + 5x + 4, exactly the polynomial that began the worksheet.

The growing powers explain the method. The leading two receives three multiplications by x before the work is finished. Negative three enters one stage later and receives two. Five receives one multiplication, while four enters at the final addition and receives none. Each coefficient therefore acquires the power assigned to its place in the original list.

Check forty-six by direct substitution

Use the original expanded polynomial for a separate arithmetic check. At x = 3, the cube is twenty-seven and the square is nine. The terms become two times twenty-seven, negative three times nine, five times three and four. Their values are fifty-four, negative twenty-seven, fifteen and four.

Adding those terms gives 54 − 27 + 15 + 4 = 46. The agreement confirms this worked calculation from two equivalent arrangements. Keep both the original expression and the nested working: checking only a second copy of the same running-value list could reproduce an unnoticed transcription error in the coefficients.

If the two answers disagree, compare inputs and signs before changing the final total. Confirm that both calculations use three, that the square coefficient is negative three and that the constant is four. Then inspect one update at a time. The first point where corresponding values disagree usually narrows the arithmetic that needs attention in this specific worksheet.

Keep a zero for a missing power

Now define a different polynomial Q(x) = 2x³ − 5x + 1 and choose x = 2. The square term is absent, so its coefficient is zero. The descending coefficient list is therefore 2, 0, −5, 1. There are four entries because the list covers powers three, two, one and zero.

Begin with two. Multiply by the input two and add the zero coefficient, leaving four. Multiply four by two and add negative five, giving three. Multiply three by two and add one, giving seven. The running-value list is 2, 4, 3, 7. The zero addition changed no value at its own stage, but that stage still included a multiplication.

Direct substitution confirms the result: two times eight minus five times two plus one equals sixteen minus ten plus one, or seven. In nested form, the complete expression is ((2 × 2 + 0) × 2 − 5) × 2 + 1. Retaining the zero makes the correspondence with all four power positions explicit.

Omitting zero would produce the list 2, −5, 1. At input two, its updates would be two, negative one and negative one. That shortened list evaluates 2x² − 5x + 1, a different polynomial whose value at two is −1. The omitted stage removed a multiplication by x and lowered the leading power.

Use the same input at every stage

Return to P(x) = 2x³ − 3x² + 5x + 4 and set x = −1. Start with two. The first update is 2 × (−1) − 3 = −5. The next is (−5) × (−1) + 5 = 10. The final update is 10 × (−1) + 4 = −6.

Parentheses around negative one keep its sign attached to the input in each line. The direct expression also gives −2 − 3 − 5 + 4 = −6. The positive square of negative one and its negative cube are accounted for automatically by the successive multiplications, while the coefficient signs remain those of the original polynomial.

At x = 0, the running values are 2, −3, 5 and 4. Each multiplication by zero removes the previous running value before the next coefficient is added. The final value is the constant four, agreeing with direct substitution. This check also exposes an extra multiplication after the constant: multiplying the finished four by zero would change the answer incorrectly.

Check order before repairing a total

Reversing the main coefficient list gives 4, 5, −3, 2. At input three, its running values become four, seventeen, forty-eight and one hundred forty-six. That is a consistent evaluation of 4x³ + 5x² − 3x + 2. It is not the original P(3). Copy the coefficient list from the stated powers before deciding that a different final number proves an arithmetic mistake.

Reversing the operations within a stage also changes the expression. At the first stage, adding negative three to the running two gives negative one, and multiplying that by three gives negative three. The required first update instead multiplies two by three before adding negative three, giving positive three. Parentheses and a complete equation for each stage make this difference visible. Once the update rule is written, use that same multiplication-then-addition order throughout the worksheet rather than choosing whichever order happens to produce a familiar number.

Explain the general prefix rule

For a coefficient list c₀ through cₙ, let the list correspond to powers n down to zero. Start with c₀. After the first update, the running expression is c₀x + c₁. After the second, it is c₀x² + c₁x + c₂. Each update multiplies the entire expression already built by x, then appends the next coefficient.

More generally, after j updates the expression is c₀xʲ + c₁xʲ⁻¹ + the continuing terms through cⱼ. Multiplying it by x raises every existing power by one. Adding cⱼ₊₁ supplies the next constant term of that prefix. This shows why the same structure continues through the full list, where the last expression is the original polynomial.

A list containing only a constant needs no updates: its starting value is already the result. A longer list needs one update for each coefficient after the first. This counts the operations defined in the worksheet; it supplies no claim about the speed, accuracy or behavior of a particular calculator, program or physical device.

Separate evaluation from neighboring algebra tasks

The equation-solving guide checks which unknown candidate satisfies a stated equation and its conditions. This worksheet instead receives x in advance and computes a polynomial's value. A resulting number is an evaluation result, without a claim that the input solves a separate equation.

The repeating-decimal fraction tutorial aligns infinite tails before subtraction. Here the coefficient list is finite and each stage preserves a polynomial prefix. The zero placeholder maintains a missing power's position rather than the width of a repeating decimal block.

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Save the inputs beside the result

Retain the original polynomial, descending coefficient list and chosen input together. For the main example, record 2, −3, 5, 4 at x = 3, the running values 2, 3, 14, 46 and the direct substitution total forty-six. For Q(2), keep the zero coefficient visible. These records let another reader reconstruct the same evaluation and see whether a proposed change alters the input, the polynomial or only an arithmetic step.